One cubic meter of nitrogen at 40°C and 340kPa is compressed isoentropically to 0.2m^3. Calculate the final pressure when the ni
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Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
Answer:
COP = 3.828
W' = 39.18 Kw
Explanation:
From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.
h1 = 238.43 KJ/Kg
s1 = 0.94575 KJ/Kg.K
From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.
h3 = h4 = hf = 95.47 KJ/Kg
For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;
h2 = 275.75 KJ/Kg
The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.
W' = m'(h2 - h1)
W' = Q'_L((h2 - h1)/(h1 - h4))
Where Q'_L = 150 kW
Plugging in the relevant values, we have;
W' = 150((275.75 - 238.43)/(238.43 - 95.47))
W' = 39.18 Kw
Formula foe COP is;
COP = Q'_L/W'
COP = 150/39.18
COP = 3.828
Answer: Resultant elevation will be 30.98 ft
